Question

Plz solve for parts a &b

A block of wood has mass of 3.67 kg and a density of 600 kg/m3. It is to be loaded with lead so that it will float in water with 0.90 of its volume immersed 5. What mass of lead is needed: a) if the lead is on the top of the wood? b) if the lead is on the bottom of the wood?

Answer 1

here,

5)

mass of wood , m1 = 3.67 kg

density of wood , p1 = 600 kg/m^3

a)

when lead is placed at the top of wood box

only wood box is immerssed in the water

let the mass of lead needed be m2

equating the forces

buoyant force = (m1 + m2 ) * g

pw * (0.9 * Volume of wood box) * g = (m1+ m2 ) * g

pw * (0.9 * m1 /p1) * g = (m1+ m2 ) * g

1000 * ( 0.9 * ( 3.67/600)) = ( 3.67 + m2)

m2 = 1.84 kg

the mass of lead is 1.84 kg

b)

density of lead , p2 = 11340 kg/m^3

when lead is placed at the bottom of wood box

wood box and lead are immerssed in the water

let the mass of lead needed be m2

equating the forces

buoyant force = (m1 + m2 ) * g

pw * (0.9 * Volume of wood box) * g + pw * volume of lead * g = (m1+ m2 ) * g

pw * (0.9 * m1 /p1) * g + pw * (m2/p2) * g = (m1+ m2 ) * g

1000 * ( 0.9 * ( 3.67/600) + m2 /11340) = ( 3.67 + m2)

m2 = 2.01 kg

the mass of lead is 2.01 kg